Once again, we will consider \(\eta = 1.\)  We will show that for Lebesgue a.e. \(E,\)  and a.e. \(\theta ,\)  the equation \(Hu = Eu\)  has solutions with appropriate decay properties.
Fix \(\theta _0 = 0,\)  and let \(H = H_{\theta _0}.\)  For each \(m \in  \mathbb {Z} ^+,\)  let \(G_m\)  be the operator on \(l^2( \mathbb {Z} ^+)\)  given by
\(\left\langle \delta _i, G_m\delta _j \right\rangle  = \delta _{i,m - 1}\delta _{j,m} + \delta _{i-1,m}\delta _{j, m-1}.\) 
For each \(k \in Z^+,\)  define new operators \(H_k^{\prime } = H - G_{L_k}\)  and \(\hat{H}_k = H - G_{L_k} - G_{L_k + 1},\)  and for every \(z \in \)  define the resolvent operators \(G(z) = (H - z)^{-1}, G_k^{\prime }(z) = (H_k^{\prime } - z)^{-1},\)  and \(\hat{G}_k(z) = (\hat{H}_k - z)^{-1}.\)  Moreover, for \(i,j \in  \mathbb {Z} ^+,\)  let the corresponding Green's functions be given by
\(G(i,j,z) &= \left\langle \delta _i, G(z)\delta _j \right\rangle \\G_k^{\prime }(i,j,z) &= \left\langle \delta _i, G_k^{\prime }(z)\delta _j \right\rangle \\\hat{G}_k(i,j,z) &= \left\langle \delta _i, \hat{G}_k(z)\delta _j \right\rangle .\) 
Now considering some \(n > L_k,\)  we can use the resolvent identity \(G(z) = G^{\prime }_k(z) - G(z) G_{L_k} G^{\prime }_k(z)\)  to obtain
\(G(1,n,z) = - G(1, L_k - 1, z)G_k^{\prime }(L_k, n, z).\) 
A similar computation with \(G^{\prime }_k(z),\)  yields the identity
\(G^{\prime }_k(z) = \hat{G}_k(z) - \hat{G}_k(z) G_{Lk + 1} G_k^{\prime }(z) ,\) 
so we have
\(G_k^{\prime }(Lk, n ,z) = -\hat{G}_k(L_k, L_k, z) G_k^{\prime }(L_k + 1, n, z) = \frac{-1}{V(L_k) - z} G_k^{\prime }(L_k + 1, n , z).\) 
Together, (REF ) and (REF ) yield
\(G(1,n,z) = G(1, L_k - 1, z) G_k^{\prime }(L_k + 1, n , z) \frac{1}{V(L_k) - z}.\) 
Whenever \(z = E + i\epsilon , \epsilon > 0,\)  we see that \(G(i,j,z)\)  and \(G^{\prime }_k(i , j , z)\)  have the form (REF ) and thus are Borel transforms of signed measures. We know (see e.g. [1]} for details) that Borel transforms have finite non-tangential limits a.e. on the real axis: \(|G(i,j,E)| = |G(i,j, E + i0)| < + \infty \)  and \(|G^{\prime }_k(i,j,E)| = |G^{\prime }_k(i,j,E + i0)| < + \infty .\)
Let us also recall Boole's equality for Borel transforms of singular measures: if \(F(z)\)  is the Borel transform of a singular measure on \( \mathbb {R} \)  such that \(\mu ( \mathbb {R} ) = 1,\)  then for any \(\lambda > 0,\)  we have \(|\left\lbrace  E: f(E) > \lambda  \right\rbrace | = 2/\lambda .\)  Since we have already shown that the spectral measures of \(H\)  for any vector \(\delta _i\)  are singular (Theorem REF  (ii) above), we conclude that
\(\left|\left\lbrace  E: |G(i,j,E)| > \lambda  \right\rbrace \right| &\le 4/\lambda ,\\\left|\left\lbrace  E: |G_k^{\prime }(i,j,E)| > \lambda  \right\rbrace \right| &\le 4/\lambda .\) 

From this, we deduce that for any \(j \ge 1, \gamma > 0,\)  \(k > 1, n > L_k\)  and \(E\in (-2,2),\)
\(\left|\left\lbrace  E: |G(1,n,E)| > \frac{f^j(L_k)^\gamma }{V(L_k) - 2} \right\rbrace \right| \le \frac{8}{f^j(L_k)^{\gamma /2}},\) 
where \(f(x) = \beta ^{-1}(x)\)  and \(f^j(x)\)  is the \(j\) -fold composition of \(f\)  with itself. Let us now fix \(j \ge 1.\)  By our choice of \(L_k,\)  \(\sum _{k = 2}^\infty (f^j(L_k))^{-\gamma /2} < \infty \)  for every \(j \ge 1\)  and \(\gamma > 0.\)  By the Borel-Cantelli lemma, for Lebesgue a.e. \(E\in (-2,2),\)  there exists a \(K(E, j)\)  such that for any \(k > K(E, j)\)  and \(n = L_k + 1\)  or \(L_k + 2,\)
\(|G(1,n,e)| \le \frac{f^j(L_k)^\gamma }{V(L_k) - 2}.\) 
Now, if \(E\)  is such that the sequence \(u_n = \left\lbrace  G(1,n,E) \right\rbrace _{n = 1}^\infty \)  exists, it necessarily solves the equation \(Hu = Eu\)  for \(n > 2.\)  Thus for any \(L_k + 2 < n \le L_{k + 1},\)  we can recover \(G(1,n,E)\)  using \(G(1,L_k + 1, E)\)  and \(G(1, L_k + 2, E)\)  and the action of the free transfer matrix:
\(\begin{pmatrix}G(1, n + 1, E)\\G(1, n , E)\end{pmatrix}=\Phi _{L_k + 2, n}(E)\begin{pmatrix}G(1, L_k + 2, E) \\G(1, L_k + 1, E)\end{pmatrix}.\) 
Since we know that the free transfer matrix is bounded, we have \( \left|\left| \Phi _{L_k + 2, n (E)} \right|\right|  \le C(E)\)  for \(L_k + 2 < n < L_{k + 1}\)  and \(E\in (-2,2).\)  Hence for \(L_k < n \le L_{k + 1}, \)  (REF ) holds for the same full measure set of \(E\)  as above and \(k > K(E).\)
It now follows that for Lebesgue a.e. \(E \in (-2,2),\)  there exists a solution \(v\)  of \(Hu = Eu\)  with \(|v(0)|^2 + |v(1)|^2 = 1\)  and a constant \(C = C(E),\)  such that for sufficiently large \(k\)  and \(n > L_k,\)
\(|v(n)| < C \frac{f^j(L_k)^\gamma }{V(L_k)} = C f^j(L_k)^\gamma \beta (L_k)^{-1} = Cf^j(L_k) L_{k + 1}^{-1/k}.\) 
Moreover, since there can be at most one subordinate solution of \(Hu = Eu\)  with the normalization property \(|v(0)|^2 + |v(1)|^2 = 1\)  which is decaying, \(v\)  must be the unique subordinate solution of \(Hu = Eu.\)  We also have, for \(m \in  \mathbb {Z} ^+, L_n < m \le L_{n + 1}\)  with \(n\)  sufficiently large,
\( \left|\left| v \right|\right| _m^2 &= \sum _{j = 1}^m |v(j)|^2 \\&= \sum _{j = 1}^{L_{k(E)}} |v(j)|^2 + \sum _{j = L_{k(E)} + 1}^m |v(j)|^2\\&\le C(E, v) + C\sum _{i = k(E)}^{n} L_{i + 1} f^j(L_i)^{2\gamma } L_{i + 1}^{-2/i}\\&\le C(E,v) + C f^j(L_n)^{2\gamma } L_{n + 1}.\) 
Now we return to considering \(H_\theta ,\)  where \(\theta \)  can vary. Recall that we can view \(H_\theta \)  as \(H\)  along with an appropriate rank-one perturbation at the origin. By the theory of rank-one perturbations (again, we refer readers to [1]} for full details), it is known that for any set \(A\subset  \mathbb {R} \)  with \(|A| = 0,\)  we have \(\mu (A) = 0\)  for Lebesgue a.e. boundary phase \(\theta .\)  Since the set of energies for with the solution \(v\)  above does not exist is a Lebesgue null set, we can conclude that for a.e. boundary phase \(\theta ,\)  the associate spectral measure \(\mu \)  is supported on the set of \(E\)  where the solution \(v\)  above exists. Furthermore, since \(\mu \)  must also be supported on the set of energies for which \(u_1\)  is subordinate, it follows that for a.e. \(\theta \)  and a.e. \(E\)  with respect to \(\mu ,\)  \(u_1\)  must coincide with \(v\)  above.
For this \(u_1\)  and \(m = L_n + L_n L_{n + 1}^{2/n},\)  we have
\( \left|\left| u_1 \right|\right| _m^2 &= \sum _{j = 1}^{L_n} |u_1(j)|^2 + \sum _{L_n + 1}^m |u_1(j)|^2\\&\le C(E,v) + C f^j(L_{n-1})^{2\gamma } L_{n} + C (m - L_n) f^j(L_n)^{2\gamma } L_{n + 1}^{-2/n} \\& = C(E,v) + C f^j(L_{n-1})^{2\gamma } L_{n} + C f^j(L_n) L_n \\&\le C( 1 + f^j(L_n)^{2\gamma } L_n).\) 
On the other hand, a similar analysis yields
\( \left|\left| u_2 \right|\right| _m^2 \ge L_n \beta (L_n)^2 \beta ^{-1}(L_n)^{-1}\) 
for \(m = L_n + L_n L_{n + 1}^{2/n}.\)
Thus, if \(g_k(x) = \beta ^{-1}(1/x^2)f^j(x),\)  and \(m = L_n + L_n L_{n + 1}^{2/n},\)  then
\(g_k(  \left|\left| u_1 \right|\right| _m^{-1}  \left|\left| u_2 \right|\right| _m^{-1})  \left|\left| u_1 \right|\right| _m^2 &\le \frac{C( 1 + f^j(L_n)^{2\gamma } L_n)}{\beta ^{-1}(L_n \beta (L_n)^2 \beta ^{-1}(L_n)^{-1})f^k(L_n \beta (L_n)^2 \beta ^{-1}(L_n)^{-1})}\\&\le \frac{C( 1 + f^j(L_n)^{2\gamma } L_n)}{2L_n f^{k-1}(L_n)}.\) 

Since this limits to 0 whenever \(k \le j,\)  and since \(j \ge 1\)  was arbitrary, we conclude that \(\dim ^+_\mathcal {F}(\mu (A\cap \cdot )) \precsim \frac{1}{\beta ^{-1}(1/t^2)}.\)
